READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, glue or correction fluid. Write your name, class and exam index number on the Answer Sheet in the spaces provided unless this has been done for you. There are thirty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C or D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate.
This document consists of 13 printed pages and 1 blank page.
12 The reaction between potassium manganate(VII) and ethanedioic acid is an example of auto-catalytic reactions, in which one of the products catalyses the reaction.
20 The following cobalt complex is known to be the functional model for biological oxygen carrier.
What is the electronic configuration of the cobalt cation in the above complex?
A [Ar] 3d6
B [Ar] 3d7
C [Ar] 3d5 4s2
D [Ar] 3d7 4s2
21 Transition metals like platinum and rhodium are found in catalytic converters fitted into cars. Which of the following properties best explains the role of transition metals in this use?
A Transition metals can exhibit variable oxidation states in their compounds as 3d and 4s electrons have similar energies.
B Transition metals form coloured ions due to absorption of energy in the visible light region to promote an electron from a lower to a higher energy 3d orbital.
C Transition metals have very high melting points because both 3d and 4s electrons are involved in forming strong metallic bond.
D Transition metals have partially filled 3d orbitals which are available for adsorption of reactant molecules.
22 What is the hybridisation of the various carbon atoms in the following molecule?
CN
OC1
C2
C4
C3
C1 C2 C3 C4 A sp2 sp3 sp sp2
B sp2 sp2 sp3 sp
C sp2 sp2 sp sp3
D sp3 sp2 sp3 sp
11
Jurong Junior College 9729/01/J2 PRELIMINARY EXAM/2018 [Turn over
23
How many stereoisomers does the above molecule have?
A 2 B 4
C 6 D 8
24 When heated with chlorine, the hydrocarbon 2,2-dimethylbutane undergoes free radical
substitution.
In a propagation step, the free alkyl radical R• is formed.
CH3CH2 C CH3
CH3
CH3
+ Cl X + HCl
How many different forms of R• are possible?
A 1 B 2
C 3 D 4
25 Which one of the following statements regarding organic reaction mechanisms is correct?
1 Electrophilic addition of alkenes involves a carbocation intermediate.
2 Electrophilic substitution of benzene involves an intermediate with 4 π electrons.
3 Nucleophilic substitution of tertiary halogenoalkanes involves a carbocation intermediate.
4 Nucleophilic addition of a carbonyl compound involves an alkoxide intermediate.
JURONG JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION 2018
CANDIDATE NAME
CLASS 18S EXAM INDEX
CHEMISTRY Higher 2
9729/02
Paper 2 Structured Questions
29 August 2018
2 hours
Candidates answer on the Question Paper.
Additional Materials: Data Booklet
READ THESE INSTRUCTIONS FIRST Write your name, class and exam index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs. Do not use staples, paper clips, glue or correction fluid. Answer all questions in the spaces provided on the Question Paper. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided.
At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
1 Nickel is an important transition metal used in the manufacture of stainless steel alloys. It is first isolated from the mineral ore niccolite by Swedish chemist Axel Fredrik Cronstedt in 1751.
For Examiner’s
Use
(a) Nickel exists naturally as a mixture of five stable isotopes, each with their own relative isotopic mass. The information about four of these isotopes is given.
isotope percentage abundance
58Ni 68.1
61Ni 1.14
62Ni 3.63
64Ni 0.93
The relative atomic mass of nickel is 58.7.
Calculate the relative isotopic mass of the fifth isotope of nickel.
Give your answer to the nearest whole number.
[2]
(b) Nickel can form complexes with H2O ligands.
In a [Ni(H2O)6]2+ complex, the presence of the H2O ligands causes the d orbitals to split into two groups at different energy levels.
(i) On the diagram below, tick the box under each of the orbitals of the higher energy level.
(d) The formula of a complex can be determined using colorimetry.
In colorimetry, light of a certain wavelength is passed through a complex ion solution. The absorbance of the light is proportional to the intensity of the colour of the solution. The more concentrated the complex ion solution, the more intense its colour and so the higher the absorbance.
An experiment is carried out to determine the formula of the complex formed between nickel and ligand X− having the structure given in (c).
The following graph was obtained when the colour intensities of mixtures of a 3 × 10−3 mol dm−3 solution of nickel(II) chloride and a 4 × 10−3 mol dm−3 solution containing X− were measured using a colorimeter at room temperature.
For Examiner’s
Use
Determine the stoichiometry of the complex and suggest its structural formula.
2 (a) Benzenediazonium chloride, C6H5N2Cl, decomposes at 50 °C and 101 kPa, according to the equation below.
C6H5N2Cl(aq) → C6H5Cl(aq) + N2(g)
The progress of the decomposition reaction of a 500 cm3 sample of C6H5N2Cl(aq) solution was studied by measuring the volume of gas produced over time.
The volume of gas produced, V, after time, t, is proportional to the amount of benzenediazonium chloride that has decomposed. The final volume of gas produced, Vfinal, is 14.6 cm3 and it is directly proportional to the original concentration of benzenediazonium chloride. The results are recorded in the graph below.
(b) Benzenediazonium chloride is an important intermediate for the production of dyes. The following scheme shows the synthesis of benzenediazonium chloride from benzene.
For Examiner’s
Use
(i) State the type of reaction and suggest the reagents and conditions for reaction I.
type of reaction: ……………………………………………….
reagents: ……………………………………………………….
conditions: …………………………………………………….. [3]
In reaction II, nitrobenzene is reduced to phenylamine via two steps.
During step 1, granulated Sn and concentrated HCl are added to nitrobenzene and the mixture is heated under reflux in a hot water bath for about half an hour. Sn, which acts as the reducing agent, is converted to Sn4+ ions.
(ii) Balance the following half-equation for the reduction of nitrobenzene in acid solution to give C6H5NH3
3 Metals have properties that make them well suited to serve as battery anodes.
They are easily oxidised from their metallic state to produce ions and electrons, where the electrons liberated are conducted throughout the external circuit. The fact that metals are physically strong and easily fashioned into any desired shape adds to their attractiveness as anodes.
Metals commonly employed as anodes in commercial batteries are listed in Table 3.1. The tabulated properties give clues as to the ability of each metal to play this role.
Metal relative atomic mass
density / g cm−3
standard electrode potential, E,
/ V
electrochemical capacity / A h g−1
lithium 6.9 0.53 −3.04 3.86
sodium 23.0 0.97 −2.71 1.17
magnesium 24.3 1.74 −2.38 2.21
iron 55.8 7.86 −0.44 0.96
zinc 65.4 7.14 −0.76 0.82
lead 207.2 11.3 −0.13 0.26
Table 3.1
For Examiner’s
Use
(a) Suggest two reasons why lithium is the best choice among the metals in Table 3.1 to be used as a battery anode.
(b) A schematic diagram of a lithium-ion battery is shown below. Lithium is the anode whereas a paste of iron disulfide (FeS2) powder mixed with powdered graphite serves as the cathode.
Figure 3.1
For Examiner’s
Use
(i) On Figure 3.1, indicate clearly the direction of electron flow. [1]
(ii) Most batteries use aqueous solutions of ionic salts or ionisable molecules as electrolytes. However, the electrolyte used in this lithium-ion battery must be non-aqueous such as dissolving LiClO4 in dimethylsulfoxide (DMSO), which is an organic solvent. Suggest why.
(c) Like enthalpy, Gibbs free energy G is a state function. Thus, Hess’ Law can likewise be applied to calculate the standard Gibbs free energy change of a reaction, ΔG,, from relevant data such as the standard Gibbs free energy changes of formation.
Table 3.2 below lists the standard Gibbs free energy change of formation, ΔGf,,
for some compounds.
species ΔGf, / kJ mol−1
FeS2(s) −160
Li2S(s) −439
Table 3.2
For Examiner’s
Use
(i) Use the data in Table 3.2 above to show that the standard Gibbs free energy change, ΔGr
,, of the overall cell reaction in the lithium-ion battery is −718 kJ mol−1.
4Li(s) + FeS2(s) → Fe(s) + 2Li2S(s)
[1]
(ii) Use the ΔGr, value given in (c)(i) to calculate the E,
(iii) Use your answer in (c)(ii) and relevant E, value in Table 3.1 to calculate the standard electrode potential of the FeS2(s)/Fe(s) half-cell at the cathode.
For Examiner’s
Use
[1]
(iv) By using one of the phrases more positive, more negative or no change, deduce the effect of increasing [Li+] on the electrode potential of
• the left-hand electrode (anode) ………………………………………
• The right-hand electrode (cathode) …………………………………. [2]
(v) Hence deduce whether the overall E,cell is likely to increase, decrease or
4 (a) Figure 4.1 is an incomplete sketch showing the melting points of some of the elements of the Period 3 (sodium to argon). Estimate and indicate, on Figure 4.1, the melting points of the four other elements: Mg, Al, S and Cl.
Figure 4.1 [2]
For Examiner’s
Use
(b) Some reactions of magnesium and its compounds are shown below.
Mg(s)
excessO2(g)
D(s) E(aq)
add waterthen filter
MgBr2(s)
pass Br2
vapour overheated Mg add X2(aq)
then hexanehexane turns orange red
waterMgBr2(aq)
(i) Identify compounds D and E.
D: ………………….. E: ……………………. [2]
(ii) Use appropriate data in the Data Booklet to deduce whether X2 is Cl2 or I2.
(c) Sodium reacts with water to form aqueous sodium hydroxide.
Na(s) + H2O(l) → Na+(aq) + OH−(aq) + ½H2(g) ΔHr
On the grid below, draw an energy cycle which can be used to calculate ΔHr, by incorporating the enthalpy changes in Table 4.1 and any relevant data from the Data Booklet.
(b) Figure 5.1 shows a reaction scheme involving 4-bromobutanone.
Compounds J and K have the following properties: • Effervescence is seen when reacted with sodium metal. • No yellow precipitate is formed when mixed with alkaline aqueous iodine. • A pale cream precipitate slowly forms when excess HNO3(aq) is added,
(b) The pentapeptide was broken down by enzymes to form shorter peptides and individual amino acids. One of the peptides is a dipeptide with the sequence gln-tyr.
For Examiner’s
Use
(i) Draw the structure of the dipeptide at pH 12.
[2]
A mixture of this dipeptide (gln-tyr) and its two constituent amino acids (gln and tyr) was subjected to electrophoresis in a buffer at pH = 12. At the end of the experiment, the following results were seen. Spots R and S remained very close together.
The three spots are due to the three species gln, tyr and gln-tyr.
(ii) Which species is responsible for spot P? Explain your answer
spot P: …………………….
explanation ………………………………………………………………………
……………………………………………………………………………………..
……………………………………………………………………………………..
…………………………………………………………………………………….. [2]
(iii) Suggest why the other two species give spots R and S that are so close together.
observations: Tyr decolourises orange Br2(aq) and forms a white precipitate while gln will not.
OR
test: neutral FeCl3(aq)
observations: Tyr will give a violet colouration while glu will not.
OR
test: NaOH(aq), heat
observations: Gln will give NH3 gas which turns damp red litmus paper blue while tyr will not.
(d) (i) sp2
(ii) Nb can act as a Bronsted base.
This is because the lone pair of electron on Na is in an unhybridised p orbital which is parallel to the adjacent π electron systems so it is delocalised into C=C and C=Nb due to p-p orbital overlap. Hence, the lone pair of electrons is not available for protonation.
However, the lone pair (of electron) on Nb is in a sp2 orbital which is on the same plane as the ring (or not parallel to adjacent π electron system) and hence it will not be delocalised into the adjacent C=C. Thus, the lone pair of electron is available for protonation.
JURONG JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION 2018
CANDIDATE NAME
CLASS 18S EXAM INDEX
CHEMISTRY Higher 2
9729/03
Paper 3 Free Response
11 September 2018
2 hours
Candidates answer on separate paper.
Additional Materials: Answer Paper
Data Booklet
READ THESE INSTRUCTIONS FIRST Write your name, class and exam index number on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a HB pencil for any diagrams, graphs. Do not use staples, paper clips, glue or correction fluid. Section A Answer all questions. Section B Answer one question. The use of an approved scientific calculator is expected, where appropriate. A Data Booklet is provided. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.
This document consists of 11 printed pages and 1 blank page.
The pKa values for stage 1 and stage 2 for some dicarboxylic acids are listed below.
n in HO2C(CH2)nCO2H pKa(1) for stage 1 pKa(2) for stage 2
1 2.83 5.69
2 4.16 5.61
3 4.31 5.41
For comparison, the pKa value of ethanoic acid, CH3CO2H, is 4.76.
(i) With reference to the table given, suggest why the pKa(1) values
• are all smaller than the pKa of ethanoic acid,
• become larger as n increases. [3]
(ii) Suggest why all the pKa(2) values in the table above are larger than the pKa of ethanoic acid. [1]
(iii) Calculate the pH of a 0.10 mol dm–3 solution of HO2CCH2CO2H. Ignore the effect of pKa(2) on the pH. [1]
(iv) Sketch the pH-volume added curve you would expect to obtain when 20 cm3 of 0.10 mol dm–3 NaOH is added to 10 cm3 of 0.10 mol dm–3 HO2CCH2CO2H.
Mark clearly the initial pH and the point(s) of maximum buffering capacity. Indicate the volume at equivalence point(s). [3]
(v) The monosodium salts of edible dicarboxylic acids can act as buffers.
Write two equations to show how monosodium butanedioate, HO2C(CH2)2CO2Na, acts as a buffer. [2]
(vi) A solution containing both HO2C(CH2)2CO2Na and NaO2C(CH2)2CO2Na forms a buffer solution. The following equilibrium is present in the solution.
HO2C(CH2)2CO2– (aq) = –O2C(CH2)2CO2
– (aq) + H+ (aq)
By choosing the correct pKa value given in the table, calculate the pH of a buffer solution made by mixing 100 cm3 of 0.5 mol dm–3 HO2C(CH2)2CO2Na and 50 cm3 of 0.3 mol dm–3 NaO2C(CH2)2CO2Na. [2]
2 Sulfur is the second member of Group 16 in the Periodic Table.
(a) Explain why sulfur has a lower first ionisation energy than phosphorus. [1]
(b) Natural gas, consisting primarily of methane, is available in large amounts and is used as a fuel. However, direct conversion of alkanes such as methane into useful products is challenging owing to their unreactivity.
(i) Give a reason why alkanes are unreactive. [1]
In 2013, scientists discovered that they could selectively convert methane to alkenes using gaseous sulfur, S2, over a suitable catalyst. This conversion takes place at over 1000 K, where sulfur exists as S2(g), as shown in reaction 1.
(ii) reaction 1 2CH4(g) + S2(g) → C2H4(g) + 2H2S(g)
Calculate the enthalpy change of reaction 1 by drawing an energy cycle using the following information.
(c) The key stage in the manufacture of sulfuric acid is the reaction between sulfur dioxide and oxygen over a catalyst.
2SO2(g) + O2(g) = 2SO3(g)
(i) When an equimolar mixture of SO2 and O2 is passed over a catalyst at T °C at an initial total pressure of 200 kPa, the percentage conversion of SO2(g) is 98%.
Calculate the equilibrium partial pressure of each of the three gases and hence, the value of Kp at temperature T °C. [3]
(ii) The table below shows values of Kp for this equilibrium at different temperatures.
temperature / °C Kp / kPa−1
25 4.0 × 1022
200 2.5 × 108
800 1.3 × 10−3
Using relevant data from the given table, explain in terms of the position of equilibrium,
Use this information to construct a fully-labelled reaction pathway diagram for the catalysed oxidation of SO2 to SO3. Indicate on your diagram, the ‘reactants’, ‘products’, ΔH1
,, ΔH2, and the activation energy, Ea, of the reaction. [3]
(f) Oxides of nitrogen are also used in other applications.
One such oxide is N2O5, which exists as [NO2+][NO3
−] in solution.
When benzene is added to a solution of N2O5 in CCl4, an excellent yield of nitrobenzene is obtained in the absence of any added catalyst.
(i) Write a balanced equation for the reaction of benzene and N2O5. [1]
(ii) Suggest a mechanism for this reaction, showing all charges and using curly arrows to show the movement of electron pairs. [2]
(iii) Compounds H has the molecular formula, C9H12. Mononitration of H produces only one organic product. Suggest the structure of H. [1]
3 (a) Copper has the highest electrical conductivity rating among the non-precious metals. The high conductivity of copper is a consequence of its electronic configuration.
(i) State the electronic configuration of Cu atom. [1]
(ii) Explain why copper is regarded as a transition element. [1]
(iii) Suggest why copper is not usually used for overhead electrical cables despite its superior conductivity. [1]
(b) Electrical wiring is the most important market for the copper industry. To be used in wiring, copper must be at least 99.9% pure. Chalcopyrite, CuFeS2, is the most common copper ore being used to obtain pure copper in a two-step process.
The first step occurs in a furnace where chalcopyrite is heated strongly with silica, SiO2, and air. The furnace reduces the copper(II) in chalcopyrite to copper.
The reaction occurring in the blast furnace can be represented by the following unbalanced equation.
CuFeS2 + SiO2 + O2 → Cu + FeSiO3 + SO2
The copper obtained is nowhere near 99.9% pure as it contains nickel and silver as minor impurities to form an alloy.
The second step occurs in an electrolytic cell where the alloy undergoes electrolysis to produce copper at or above 99.9% purity.
(i) Identify the element that undergoes oxidation in the furnace and state the initial and final oxidation numbers of this element. [1]
(ii) In the furnace, the element O is reduced alongside with the element Cu. Using oxidation numbers or otherwise, write a balanced equation for the reaction occurring in the furnace. [1]
(iii) Draw a fully labelled diagram of the electrolytic cell where copper purification occurs. [2]
(iv) Explain, by quoting relevant E, values, what happens to the nickel and silver impurities during this purification procedure. [4]
(c) When dilute aqueous ammonia was added to copper(II) sulfate solution, a pale blue precipitate was first observed which dissolved to give a dark blue solution when aqueous ammonia was added in excess. In these reactions, the ammonia can act as a Bronsted-Lowry base and as a Lewis base.
Illustrate the meaning of Bronsted-Lowry base and Lewis base using the reactions of aqueous copper(II) ions with ammonia. Write equations to account for the observations noted in these reactions. [4]
(d) In 2013, scientists suggested that too much copper in our diet may contribute to Alzheimer’s disease, a neurological disorder in which death of brain cells causes memory loss and cognitive decline. In patients with Alzheimer’s disease, the concentration of acetylcholine, a neurotransmitter essential for processing memory and learning, is found to be lower.
The structure of acetylcholine is shown below.
C C
H
H
H
O
O C C
H
H
N
H
H CH3
CH3
CH3
+
When acetylcholine undergoes hydrolysis in the presence of an enzyme, two products are formed; one of which is a charged species called choline.
(i) Draw the structures of these two products. [2]
(ii) As a precursor to the synthesis of acetylcholine, choline is an essential nutrient in our diet. Choline can be produced in a sealed reaction vessel by the reaction of anhydrous trimethylamine, N(CH3)3, with epoxyethane and HCl under moderate pressure.
Copy and complete the diagram below to suggest a mechanism to show how choline is formed. Show all charges and relevant lone pairs and show the movement of electron pairs by using curly arrows.
4 (a) Thionyl chloride, SOCl2, is a colourless liquid that is primarily used in the chlorination of organic compounds.
(i) Draw the shape of SOCl2. [1]
(ii) Explain which molecule, SOCl2 or COCl2, is expected to have a larger bond angle. [1]
(b) When SOCl2 is reacted with a carboxylic acid to produce an acyl chloride, two acidic gases are formed.
SOCl2(l) + RCO2H → RCOCl + SO2(g) + HCl (g)
A 1.00 g sample of a carboxylic acid RCO2H was treated in this way, and the gases were absorbed in 60.0 cm3 of 0.500 mol dm–3 NaOH(aq). The excess NaOH was titrated with 0.500 mol dm–3 HCl(aq). It required 10.8 cm3 of the HCl(aq) solution to reach the end-point.
(i) Write equations for the complete reactions between
• NaOH and HCl,
• NaOH and SO2. [2]
(ii) Calculate the total number of moles of NaOH that reacted with the acidic gases, SO2 and HCl. [1]
(iii) Calculate the number of moles of RCO2H that produced the SO2 and HCl. [1]
(iv) Hence calculate the Mr of the carboxylic acid, RCO2H. [1]
(v) The R group contains carbon and hydrogen only.
Suggest the structure of RCO2H. [1]
(c) Using SOCl2 in the initial step, carboxylic acids can be converted into primary amines by the following sequence of reactions.
RCO2H RCOCl D RCH2NH2
(i) Identify the structure of D. [1]
(ii) Suggest the reagents needed for step 2 and step 3. [2]
(iii) State the types of reaction for step 2 and step 3. [1]
5 (a) (i) Describe the reaction of SiCl4 with water and include any observations you can make. Write an equation for the reaction that occurs. [2]
(ii) Carbon is in the same group as silicon. However, CCl4 does not react with water. Explain why. [1]
(b) Chlorine-containing organic compounds can also differ in their reactivity towards hydrolysis reactions.
Compounds A and B are isomers with the molecular formula, C7H7Cl. Compound A reacts with aqueous NaOH on heating but compound B does not.
Draw the structures of compounds A and B and explain the difference in their reactivity towards aqueous NaOH. [2]
(c) Heating tin with hydrochloric acid produces hydrogen gas. Careful removal of water from the resulting product produces white solid SnCl2.
In contrast, passing chlorine gas over heated tin produces colourless liquid SnCl4 as the only product.
Using relevant E, from the Data Booklet, explain the above observations. [2]
(d) Cumin is a spice used to flavour food. One of the key organic compounds responsible for the smell of cumin is cuminyl alcohol. It can be synthesised from cumene as shown below. Aluminium chloride is used as a catalyst with reagent X in step 1.
(i) Suggest the identity of reagent X. [1]
(ii) Suggest reagents and conditions for steps 2 and 3 and draw the structure of the intermediate compound Y. [3]
5 (e) Carvone, an isomer of cuminyl alcohol, is most responsible for the flavour of caraway, dill and spearmint.
carvone
(i) When carvone is treated with NaBH4, compound M, C10H16O, is produced. Draw the structure of M and write an equation to represent this reaction. [2]
(ii) Compound N is isomeric with compound M.
On treatment with hot concentrated KMnO4, N gives two compounds, P, C3H6O, and Q, C4H4O5. P and Q are formed in the mole ratio of 2 : 1.
All three compounds, N, P and Q give an orange precipitate with 2,4-dinitrophenylhydrazine but only N forms a silver mirror with Tollens’ reagent. One mole of Q reacts with two moles of aqueous sodium hydrogencarbonate.
Suggest structures for N, P and Q and explain the observations described above. [7]
The following should be clearly marked and labelled: () Both axes are labelled with units stated for x-axis. () Curve starts at pH 1.92 (ecf) () First maximum buffering capacity at 5 cm3 and pH 2.83. () Second maximum buffering capacity at 15 cm3 and pH 5.69. () First equivalence point at
[1m] quote the correct pKa (5.61) in working [1m] pH calculation (ecf from wrong pKa)
2 (a) Inter-electronic repulsion between paired 3p electrons makes it easier to remove one of the paired 3p electrons than to remove the unpaired 3p electron from phosphorus. [1m]
(b) (i) [1m] Either one of the following:
Very strong C–C and C–H bonds / high bond energies
Non-polar / C and H have similar electronegativities
(ii) Since Kp at 25 °C is much larger than 1, the position of equilibrium in 2SO2(g) + O2(g) = 2SO3(g) lies very much to the right [1m], so the forward reaction is spontaneous and ΔG, is negative. [1m]
As Kp decreases with increasing temperature, it implies that the position of equilibrium shifts to the left [1m] with increasing temperature to absorb some heat. Hence, the backward reaction is endothermic and the forward reaction is exothermic, so ΔH has a negative sign. [1m]
×
×
a a[salt] n(salt)pH = pK + lg = pK + lg[acid] n(acid)
(not 3d9 4s2 due to the extra stability of 3d10 and the similar energies of 3d and 4s electron)
(ii) Copper is a d-block element that forms one or more stable ions with incompletely filled d-orbitals. [1m]
(iii) Copper has high density/ is too heavy for overhead use. [1m]
(b) (i) S is oxidised from –2 to +4 [1m]
(ii) 2 CuFeS2 + 2 SiO2 + 5 O2 → 2 Cu + 2 FeSiO3 + 4 SO2 [1m]
(iii)
[1m] these labels on the diagram: impure copper, pure copper, CuSO4(aq)
[1m] battery with 2 electrodes dipped in a common electrolyte; battery must show the correct polarity with respect to the impure copper.
(iv) species present:
Cu2+(aq), SO42−(aq), H2O(l), impure Cu (anode) and impurities such as Ag, Ni
O2 + 4H+ + 4e– = 2H2O E, = +1.23 V
Cu2+ + 2e– = Cu E, = +0.34 V Ni2+ + 2e– = Ni E, = –0.25 V Ag+ + e– = Ag E, = +0.80 V
[1m] quote E, values for Cu2+/Cu, Ni2+/Ni and Ag+/Ag
At the anode, Cu(s) is oxidised to Cu2+(aq) in preference over H2O as E,(Cu2+/Cu) is more negative than E,(O2/H2O). Ni(s) is also oxidised to Ni2+(aq) (which goes into the solution) as E,(Ni2+/Ni) is more negative than E,(Cu2+/Cu). Ag(s) is not oxidised to Ag+ as E,(Ag+/Ag) is more positive than E,(Cu2+/Cu). ∴Ag(s) is collected as “anode sludge”. 2H2O + 2e– = H2 + 2OH– E, = –0.83V Cu2+ + 2e– = Cu E, = +0.34 V Ni2+ + 2e– = Ni E, = –0.25 V
At the cathode, Cu2+(aq) is reduced to Cu(s) in preference over H2O as E,(Cu2+/Cu) is more positive than E,(H2O/H2).
Ni2+(aq) is not reduced to Ni(s) as E,(Ni2+/Ni) is more negative than E,(Cu2+/Cu). ∴Ni2+(aq) remains in the solution.
Note: The species is more easily oxidised when the E, is more negative (or less positive).
Note: The species is more easily reduced when the E, is more positive (or less negative).
Or [Cu(H2O)6]2+(aq) + 4NH3(aq) → [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l) dark blue soln
When added in excess, NH3 acts as a Lewis base where N in NH3 ligand donates a lone pair of electrons to Cu2+, forming strong dative bonds with Cu2+ to give the stable [Cu(NH3)4(H2O)2]2+ complex. [1m]
(d) (i)
C C
H
H
H
O
OH
[1m] and [1m]
(ii)
H2C CH2
O
H+
H2C CH2
OH+
H2C CH2
OH+
H2C C
OH+N(CH3)3
H
HN(CH3)3
H2C CH2
O
H+
+
H2C C
OH+N(CH3)3
H
H
H2C CH2
O
H
+
H2C CH2
O
H
N(CH3)3
H2C CH2
O
H+
H2C C
OH+N(CH3)3
H
HN(CH3)3
[1m] for each curly arrow with the associated lone pair and any positive charge
(ii) COCl2 will have a larger bond angle. [1m] bonus
COCl2 has lesser number/ 3 sets of electron pairs (or state 3 bp, 0 lp) which can be further apart to minimise repulsion compared to SOCl2 that has 4 electron pairs (or state 3 bp, 1 lp). [1m]
(b) (i) NaOH + HCl → NaCl + H2O [1m]
2 NaOH + SO2 → Na2SO3 + H2O [1m]
Allow if 2nd eqn is split into 2 eqns: SO2 + H2O → H2SO3 2NaOH + H2SO3 → Na2SO3 + 2H2O
Reject: NaOH + SO2 → NaHSO3 for NaOH is added in excess
(ii) n(NaOH) reacted =
60.0 10.80.500
1000
− ×
= 0.0246 mol [1m]
(iii) RCO2H : SO2 + HCl : NaOH 1 : 1 : 3
n(RCO2H) = 0.0246
3 = 0.00820 mol [1m]
(iv) Mr (RCO2H) =
1.00
0.0082 = 122 [1m] no units, 3 s.f., ecf
(v) RCO2H: mass units of –CO2H = 45
mass units of –R = 122 – 45 = 77
estimated number of C in –R = 77/12 = 6.4 ( R contains 6 carbon atoms)
U + I2/OH– → yellow ppt Iodoform test/ oxidation ()
U has the structure CH3–C=O ()
5-6(): [2m]; 3-4(): [1m]
angelic acid T U V
O
O
OH
[1m] for each structure
(ii) angelic acid : cis-trans isomerism [1m]
compound T : enantiomerism [1m]
5 (a) (i) SiCl4 reacts with water / hydrolyses completely in water to give a strongly acidic solution. [1m] bonus
A white solid/ppt of SiO2 will be observed/ white (or steamy) fumes of HCl will be observed. [1m]
SiCl4 + 2H2O → SiO2 + 4HCl [1m]
(ii) Unlike silicon, carbon does not have energetically accessible empty 3d orbitals to accept a lone pair of electrons from water molecules. [1m]
(b) CH2Cl
Cl
CH3
A B (-CH3 and Cl groups can be in any position)
[1m] for both structures of A and B
For compound B, delocalisation of the lone pair of electrons on Cl into the π-electron cloud of benzene ring imparts double bond character to the C–Cl, strengthening the C–Cl bond. Hence, B does not react with NaOH(aq). [1m]
